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**Hint:**We have been given a quadratic equation of $x$ as ${{x}^{2}}-6x+25=0$. We use the quadratic formula to solve the value of the $x$. we have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

**Complete step by step solution:**

We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of $x$ will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method. The root part $\sqrt{{{b}^{2}}-4ac}$ of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is called the discriminant of the equation.

In the given equation we have ${{x}^{2}}-6x+25=0$. The values of a, b, c is $1,-6,25$ respectively.

We put the values and get $x$ as \[x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\times 25\times 1}}{2\times 1}=\dfrac{6\pm \sqrt{-64}}{2}=\dfrac{6\pm 8i}{2}=3\pm 4i\]

The roots of the equation are imaginary numbers. So, values of x are $x=3\pm 4i$.

The discriminant value being negative square, we get the imaginary numbers root values.

In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is non-square. ${{b}^{2}}-4ac={{\left( -6 \right)}^{2}}-4\times 25\times 1=-64$.

This is a negative square value. That’s why the roots are imaginary.

**Note:**We have been given the equation ${{x}^{2}}-2x-4=0$. We form the square part in ${{x}^{2}}-6x+25$.

The square form of subtraction of two numbers be ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.

We have ${{x}^{2}}-6x+25={{x}^{2}}-2\times x\times 3+{{3}^{2}}+16$.

Forming the square, we get ${{x}^{2}}-6x+25={{\left( x-3 \right)}^{2}}+{{4}^{2}}$.

We get ${{\left( x-3 \right)}^{2}}+{{4}^{2}}=0$. Taking solution, we get

$\begin{align}

& {{\left( x-3 \right)}^{2}}+{{4}^{2}}=0 \\

& \Rightarrow {{\left( x-3 \right)}^{2}}=-{{4}^{2}} \\

& \Rightarrow \left( x-3 \right)=\pm 4i \\

& \Rightarrow x=3\pm 4i \\

\end{align}$.

Thus, the solution of the equation ${{x}^{2}}-2x-4=0$ is $x=3\pm 4i$.

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